general relativity - A question about the physics involved in tracking satellites such as those used in the GPS system

I know that besides the effects of Newton's theory of Gravitation on the satellite's motion, one has to take account of the retardation of the satellite's clocks when compared to earth-fixed clocks. Is this just an effect of Special Relativity? Or are there other effects that must be taken into account, which would not be predicted by Special Relativity alone, but which only occur in General Relativity?

Answer

The short answer is yes. An object moving at non-relativistic velocity $\vec{v}$ in a weak gravitational field will have a proper time $\Delta \tau$ elapse that is related to the time $\Delta t$ on distant clocks (far from the Earth) by the equation $$\frac{\Delta \tau}{\Delta t} \approx 1 - \left( \frac{1}{2} \frac{v^2}{c^2} + \frac{G M_E}{r c^2} \right)$$ where $r$ is the distance from the center of the Earth and $M_E$ is the Earth's mass. The first term in parentheses is just the change in the Lorentz factor $\gamma = (1 - v^2/c^2)^{-1/2}$ to lowest non-vanishing order in $v$. The second term, however, can be seen to have something to do with gravity (in fact, the general expression would be $\Phi/c^2$, where $\Phi$ is the Newtonian gravitational potential.) For a satellite in circular orbit, it's not too hard to show that $v^2 = G M_E/r$; thus, you can see that the correction due to special relativity is one-half as large as the correction due to gravity.

Of course, we don't measure the time difference relative to clocks at infinity; we measure the time difference relative to clocks on the ground, which are themselves moving (due to rotation of the Earth) and in a gravitational field. To find the true correction effects, you could combine the correction factor for orbital clocks relative to infinity with the correction for Earth-based clocks relative to infinity. This would lead to a fractional correction relative to Earth of something like $$\frac{\Delta \tau_s}{\Delta \tau_g} \approx 1 - \left( \frac{1}{2} \frac{v_s^2}{c^2} + \frac{G M_E}{r_s c^2} \right) + \left( \frac{1}{2} \frac{v_g^2}{c^2} + \frac{G M_E}{r_g c^2} \right)$$ where the subscripts $g$ and $s$ stand for "ground" and "satellite" respectively. If we assume a station at the Earth's equator, then we have $v_s \approx 464$ m/s and $r_s \approx 6380$ km; for GPS satellites, we have $r_s = 26 580$ km (an altitude of approximately 20,200 km) and, since they are in approximately circular orbits, $v_s = 3 872$ m/s (using the formula above.) Thus, the overall correction factor due to special-relativistic effects would be $$\frac{1}{2} \left( \frac{v_g^2}{c^2} - \frac{v_s^2}{c^2} \right) \approx -8.2 \times 10^{-11} \approx - 7 \, \mu\text{s / day}$$ while the correction factor due to gravitational effects would be $$\frac{G M_E}{r_g c^2} - \frac{G M_E}{r_s c^2} \approx 5.2 \times 10^{-10} \approx 46\, \mu \text{s / day}.$$ Thus, the correction effects work in opposite directions, but you can see that the gravitational effects cannot be neglected compared to the special-relativistic effects.